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As shown in the figure, in Young's double slit experiment, a thin plate of thickness $$t = 10 \mu m$$ and refractive index $$\mu = 1.2$$ is inserted infront of slit $$S_1$$. The experiment is conducted in air ($$\mu = 1$$) and uses a monochromatic light of wavelength $$\lambda = 500$$ nm. Due to the insertion of the plate, central maxima is shifted by a distance of $$x\beta_0$$. $$\beta_0$$ is the fringe-width before the insertion of the plate. The value of the $$x$$ is ______
Correct Answer: 4
$$\Delta y = \frac{D(\mu - 1)t}{d}$$
$$\beta_0 = \frac{\lambda D}{d}$$
$$\Delta y = x \beta_0 \implies \frac{D(\mu - 1)t}{d} = x \frac{\lambda D}{d}$$
$$x = \frac{(\mu - 1)t}{\lambda}$$
$$x = \frac{(1.2 - 1) \times 10 \times 10^{-6}}{500 \times 10^{-9}} = \frac{0.2 \times 10^{-5}}{5 \times 10^{-7}} = \frac{2 \times 10^{-6}}{5 \times 10^{-7}} = 4$$
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